Problem: $f(x)=\begin{cases} \sqrt{7+x}&\text{for }-7\leq x\leq-3 \\\\ x^2-5&\text{for }x>-3 \end{cases}$ Find $\lim_{x\to -3}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $2$ (Choice C) C $4$ (Choice D) D The limit doesn't exist.
Answer: $x=-3$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to -3}f(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $-3$ from the left. We will use the fact that $f(x)=\sqrt{7+x}$ for $x$ -values smaller than $-3$. $\begin{aligned} &\phantom{=}\lim_{x\to -3^-}f(x) \\\\ &=\lim_{x\to -3^-}\sqrt{7+x} \\\\ &=\sqrt{7+(-3)}&\gray{\text{Direct substitution}} \\\\ &=2 \end{aligned}$ Let's find the limit as $x$ approaches $-3$ from the right. We will use the fact that $f(x)=x^2-5$ for $x$ -values greater than $-3$. $\begin{aligned} &\phantom{=}\lim_{x\to -3^+}f(x) \\\\ &=\lim_{x\to -3^+}x^2-5 \\\\ &=(-3)^2-5&\gray{\text{Direct substitution}} \\\\ &=4 \end{aligned}$ $2\neq 4$ so the one-sided limits aren't equal. This means that the two-sided limit $\lim_{x\to -3}f(x)$ doesn't exist.